1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Questions on Trigonometric equation solving

  1. Apr 5, 2015 #1
    Hello reader,

    I have an exam really soon and it includes a good bit of trigonometry, but I'm having problems with the trig stuff because this exam does not allow calculators and since I was dependent on the calculator, I haven't memorized anything about the trigonometric functions. I don't know the basic angles trigonometry and I don't know when they are positive or negative , etc. I basically want the method to master trigonometric functions in terms of what I need to memorize (and if there's something else to study as well).

    What do I need to know for the exam?

    And another thing. This is a sample question :

    Find x ∈ [0, π/2] such that : cos^2 2x − 3 sin^2 2x = 0

    I don't even know where to start when solving this thing, and will greatly appreciate a method of solving this type of questions and maybe some other examples like this.

    Thank you in advance.
    Last edited: Apr 5, 2015
  2. jcsd
  3. Apr 5, 2015 #2


    User Avatar
    Science Advisor

    I would suggest moving (-3sin(2x)) to the other side of the equals-sign and then try to remember all the trig functions.
  4. Apr 5, 2015 #3
    "I would suggest moving (-3sin(2x)) to the other side of the equals-sign and then try to remember all the trig functions."

    Sorry, I made a mistake writing the original question, both sin and cosine were squared, as you can now see in the edited version.

    However, after moving one to the other side, I don't know what to do!
  5. Apr 5, 2015 #4


    User Avatar
    Science Advisor

    You do not remember the trig function called tangent (=sin/cos)?
  6. Apr 5, 2015 #5


    User Avatar
    Science Advisor

    Use [itex]cos^2(2x)=1-sin^2(2x)[/itex] and then solve for sin(2x).
  7. Apr 6, 2015 #6
    Sorry, I made another mistake in writing down the question, but here are my steps :

    1- take sin^2 to the other side

    2-divide by cos^2 x, the cos^2 x on one side cancels out and becomes 1 , and the sin^2 x on the other side becomes tan^2 x

    3- so now we have : 1 = 3 tan^2 (x)

    1/3 = (tan x)^2

    tan x = square root of (1/3)

    Then what?
  8. Apr 6, 2015 #7
  9. Apr 6, 2015 #8
    Ok, so it will become : x= tan^-1 [squareroot(1/3)]

    Now what do I need to memorize in order to know what tan inverse of squareroot of 1/3 is? I remember there was some sort of way to memorize these things but I don't remember what it was.
  10. Apr 6, 2015 #9
  11. Apr 9, 2015 #10
    One of the common mistakes I would make in trigonometric equations was to cancel out roots. For example, sin^4 x - sin^6 x = 0
    Can be written by taking sin^4 x common. Don't forget it to equate it to 0. I used to just cancel it and miss out a root.

    Remember the domains of all the functions. Other than that basic algebraic manipulations should give you the answer
  12. Apr 9, 2015 #11
    You could write it all out as complex exponentials and see how far you can simplify.
  13. Apr 9, 2015 #12
    Ok, the way to memorise it is pretty simple. You assign the so called standard fractions, and find their square roots to sine.
    The fractions are 0, 1/4, 1/2, 3/4, 1.

    Sine 0, 30, 45, 60 and 90 are.
    0, 1/2, sqrt(1/2), sqrt(3)/2 and 1 respectively.

    We assign the values of cos in the reverse order
    They are cos 90, 60, 45, 30 and 0.

    For tan, we divide sin by cos
    Tan 0, 30, 45, 60 and 90 are ...
    0, sqrt(1/3), 1 sqrt(3) and infinity.

    The other three trigonometric ratios are reciprocal a.

    Hope this helped.
  14. Apr 10, 2015 #13


    User Avatar
    Science Advisor

    Start with an equilateral triangle, each side of length 2, and draw a perpendicular to one side from the opposite vertex. That divides the triangle into two right triangles having angles 30, 60, 90. Further, the hypotenuse of each right triangle has length 2 and one leg, half of a side has length 1. The Pythagorean theorem shows that the length of the other leg, the perpendicular, is [itex]\sqrt{3}[/itex]. From that you can calculate the trig functions for 30 and 60 degrees.

    An isosceles triangle has acute angles of 90/2= 45 degrees. Taking each leg to be 1, the hypotenuse has length [itex]\sqrt{2}[/itex]. Use that to find all trig functions for 45 degrees.
  15. Apr 10, 2015 #14


    Staff: Mentor

    Here is how I've seen it.
    sin(0°) = ##\frac {\sqrt{0}} {\sqrt{4}}## = cos(90°)
    sin(30°) = ##\frac {\sqrt{1}} {\sqrt{4}}## = cos(60°)
    sin(45°) = ##\frac {\sqrt{2}} {\sqrt{4}}## = cos(45°)
    sin(60°) = ##\frac {\sqrt{3}} {\sqrt{4}}## = cos(30°)
    sin(90°) = ##\frac {\sqrt{4}} {\sqrt{4}}## = cos(0°)
  16. Apr 10, 2015 #15
    That's pretty neat too, Mark44
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook